alex_white101 wrote:well then, how come i attacked a 35 on 35 and ended 3 vs 25?!?!?!

he said a 50.4% chance to succeed

Well if he says the situation ended with 3 vs. 25, technically he didn't fail. He still has a possibility of winning the territory. It's all about believing =)

Believe what? That 3 troops can defeat 25?

Now I believe that am losing my mind. Can I call on some one's girlfriend please. Or some one offer me a half filled keg gar of Margarita. =)

I guess that "Gambit" site help me to lose my mind as well. I can't figure out like it reads there. Attackers = 3. Defenders = 2. Rolls = 7776! My god man! From what planet does this come from. And watch out if there is a split? Lol.

It only states the number of possible dice roll combinations with attack and defense. It is 6^5 which is indeed 7776. Check your maths, in CC they are very useful. I have created a formula to know whether I have a region correctly defended. I will only mention it involves a square root.

Dice don't even out, that's not how odds work. It all depends on the circumstances, if it's a spot that needs to be broken, you might want to go on a little longer, if it doesn't 20v25 might be the moment to stop.

mr. CD wrote:Dice don't even out, that's not how odds work. It all depends on the circumstances, if it's a spot that needs to be broken, you might want to go on a little longer, if it doesn't 20v25 might be the moment to stop.

Yes it is. Over a number of rolls, the odds of just one player winning EVERY roll diminishes greatly.

Take a coin flip, the odds that Player A wins is 50%.

But do the flip again. The odds that Player A wins TWICE in a row is only 25% (.5 * .5)

The odds that Player A wins THREE TIMES in a row is 12.5%.

Etc.

So, since rolls are similar to a coin flip (albeit more complex), you have to figure that if you are going to roll X times, the odds are extremely small that you are going to lose every single roll, and thus if you've already lost a handful of rolls, statistically you should start winning a few (that it will "even out"). To lose 11 times in a row (as the guy who went from 25 to 3 did) is not very likely at all.

I mean, that's the whole premise of this thread - that the larger the armies involved, the more likely that the attacker will win BECAUSE the dice will "even out" in favor of the best odds.

If it's 25v25 (as in our example) you can reasonably expect to lose a handful of rolls in a row but still expect to win the attack overall because, as pointed out, starting at 25v25 gives you the overall advantage (a slight advantage, but one nonethless).

So it is reasonable to assume that even if I've lost 4-5 rolls in a row, eventually I should start winning some because odds are that I'm suppose to win overall, and how else am I suppose to win unless I start winning some rolls?

When you think about it that is also the main premise in Trench Warfare settings. If you have a large stack that has to be approached, that is to say that the surrounding regions of your large stack also belong to you with only 1's on them, then your large stack has the advantage of striking the approaching force first. And so if you have more than just 11 troops then you have the advantage in the defense.

In regular settings your defensive stack would not have that advantage as the sweeping stack can cut across the game map and simply wipe out all large stacks with the 54% odds in the attackers favor. But in TW games the defender is the attacker and can even retreat (if possible) in order to maintain that attacker odds advantage in the defense of the position.

ZeekLTK wrote:So, since rolls are similar to a coin flip (albeit more complex), you have to figure that if you are going to roll X times, the odds are extremely small that you are going to lose every single roll, and thus if you've already lost a handful of rolls, statistically you should start winning a few (that it will "even out"). To lose 11 times in a row (as the guy who went from 25 to 3 did) is not very likely at all.

Actually when it come to the coin toss, I for one, can make the coin come out the same way every time or most of the time by controlling the circumstances or conditions of the toss. Catching the toss in mid-flight for example, at precisely the same interval each time and the same way each time helps. The starting position of the coin, whether it be heads up or down should also be the same each time. The pressure used on the coin by the thumb, exactly the same in each toss. This takes practice but can be achieved.

Viceroy63 wrote:Actually when it come to the coin toss, I for one, can make the coin come out the same way every time or most of the time by controlling the circumstances or conditions of the toss. Catching the toss in mid-flight for example, at precisely the same interval each time and the same way each time helps. The starting position of the coin, whether it be heads up or down should also be the same each time. The pressure used on the coin by the thumb, exactly the same in each toss. This takes practice but can be achieved.

Okay, but I was using the coin toss as an example of the dice on this site (which can't be influenced). If you are going to attack 11 times with 3v2 dice, it is not reasonable to expect to lose every single roll. Therefore, if you have lost 4-5 rolls in a row, it IS reasonable to expect to win a few of the next rolls, because odds of losing 3v2 every single time is not very high (especially the more times you roll). That was my point.

So, in the example we were using, the 25v3... if you've rolled enough to get down to say 17v3, why should you stop attacking? What are the odds that you will lose ANOTHER roll 0-2 when you've already lost 4 in a row? They are not very high, so you should keep rolling and expect to win soon. Such as the coin toss odds - if you've already lost 4 in a row, odds of losing another (5 in a row) is 3.13%, which means odds are 96.88% that you should win the next one. In the example about the 25v3, the odds of losing 22-0 (11 in a row) are 0.49% (if you assume each roll is 50/50, which they aren't, but for simplicity). Which means over 11 rolls, he had a 99.51% chance of winning at least ONE roll. Very unlucky for him, but it also means he did the right thing to keep attacking because he had such great odds of winning.

I was re-reading one of my earlier post and saw the error of my logic or my keyboard does not type my thoughts correctly. LOL. I had wrote...

Viceroy634 wrote:And so if you have more than just 11 troops then you have the advantage in the defense.

Correction: one never has the advantage in the defense no matter how many troops. But I do agree with that 11 troops vs. 11 troops, the Defender does have the very good possibility that he won't be conquered by the 11 attacking troops.

While I have personally conquered 11 troops with just a stack of 4 the percentage is probably less than 1% of the times. So for the most part the defender does usually break even or fairs better in that situation.

Viceroy63 wrote:Actually when it come to the coin toss, I for one, can make the coin come out the same way every time or most of the time by controlling the circumstances or conditions of the toss. Catching the toss in mid-flight for example, at precisely the same interval each time and the same way each time helps. The starting position of the coin, whether it be heads up or down should also be the same each time. The pressure used on the coin by the thumb, exactly the same in each toss. This takes practice but can be achieved.

Okay, but I was using the coin toss as an example of the dice on this site (which can't be influenced). If you are going to attack 11 times with 3v2 dice, it is not reasonable to expect to lose every single roll. Therefore, if you have lost 4-5 rolls in a row, it IS reasonable to expect to win a few of the next rolls, because odds of losing 3v2 every single time is not very high (especially the more times you roll). That was my point.

So, in the example we were using, the 25v3... if you've rolled enough to get down to say 17v3, why should you stop attacking? What are the odds that you will lose ANOTHER roll 0-2 when you've already lost 4 in a row? They are not very high, so you should keep rolling and expect to win soon. Such as the coin toss odds - if you've already lost 4 in a row, odds of losing another (5 in a row) is 3.13%, which means odds are 96.88% that you should win the next one. In the example about the 25v3, the odds of losing 22-0 (11 in a row) are 0.49% (if you assume each roll is 50/50, which they aren't, but for simplicity). Which means over 11 rolls, he had a 99.51% chance of winning at least ONE roll. Very unlucky for him, but it also means he did the right thing to keep attacking because he had such great odds of winning.

His logic is completely wrong. "and thus if you've already lost a handful of rolls, statistically you should start winning a few (that it will "even out")."Wrong"Such as the coin toss odds - if you've already lost 4 in a row, odds of losing another (5 in a row) is 3.13%, which means odds are 96.88% that you should win the next one. "This is more than being just wrong and shows a serious misunderstanding of maths, probabilities and gameplay. If you ve already lost 4 times in a row, chances you lose a fifth one is still 50%, not 3%

Kaskavel wrote:His logic is completely wrong. "and thus if you've already lost a handful of rolls, statistically you should start winning a few (that it will "even out")."Wrong"Such as the coin toss odds - if you've already lost 4 in a row, odds of losing another (5 in a row) is 3.13%, which means odds are 96.88% that you should win the next one. "This is more than being just wrong and shows a serious misunderstanding of maths, probabilities and gameplay. If you ve already lost 4 times in a row, chances you lose a fifth one is still 50%, not 3%

OK; But if you start out 25 troops to 3 troops, What are the odds of losing 22 troops (11 consecutive roll loses in a row) to just 3 troops?

I really don't do much complex math but I can't believe that those are odds, what ever they may be, that occur too often?

Even though I agree it could happen, but why should you stop attacking after losing three rolls in a row?

I'll grant you that the over all situation should be considered and if you need the troops then you should stop attacking the 3 especially if you go below winning odds like when you get down to just 3 or 4 troops, but while you still outnumber the 3 troops 2 to 1 and it's just another attack for a card or a region, then why stop attacking just because you lost a hand full of consecutive rolls?

Kaskavel wrote:His logic is completely wrong. "and thus if you've already lost a handful of rolls, statistically you should start winning a few (that it will "even out")."Wrong"Such as the coin toss odds - if you've already lost 4 in a row, odds of losing another (5 in a row) is 3.13%, which means odds are 96.88% that you should win the next one. "This is more than being just wrong and shows a serious misunderstanding of maths, probabilities and gameplay. If you ve already lost 4 times in a row, chances you lose a fifth one is still 50%, not 3%

OK; But if you start out 25 troops to 3 troops, What are the odds of losing 22 troops (11 consecutive roll loses in a row) to just 3 troops?

I really don't do much complex math but I can't believe that those are odds, what ever they may be, that occur too often?

Even though I agree it could happen, but why should you stop attacking after losing three rolls in a row?

I'll grant you that the over all situation should be considered and if you need the troops then you should stop attacking the 3 especially if you go below winning odds like when you get down to just 3 or 4 troops, but while you still outnumber the 3 troops 2 to 1 and it's just another attack for a card or a region, then why stop attacking just because you lost a hand full of consecutive rolls?

With math taking into account your looking at an overall number, the chance that you lose 25v3 is low. Somewhere in the thread someone posted the % of it. Though the chance you loss 11 consecutive rolls in a row is low. But if you look at the chance of every roll solo. The chance that you lose is the same very time.

Lets make this an example.

I have a coin, head or tails.

First time Head.Second time Head.Third time Head....

If it goes on to keep on being Head.

100th head.

The chance that if you coinflip 100 times and you get head 100 times is pretty low, exactly it is 1/2^100.

But what about the 101th coin flip.

The chance of this coin flip being head or tails is still 50%. Caus there is 2 options, though overall it happening that you get 101th heads is 1/2^101 this becuase your looking into it as a sequence and taking into account the 100 flips before that.

Thats what he is doing the same with the rolls, he is looking into a sequence. The sequence of losing 25v3 is a pretty low chance. Because its the chance of rolling 3 against 2 dices and that ^25. But if it in the end becomes 9v3. Its still the same chance that you win that roll instead of looking into the sequence, caus the sequence can end every time is what you take into account. So it could end on the 10th roll but also on the first. But the chance that every roll goes through with a total loss is low.

Urs

Koganosi

Last edited by Koganosi on Sat Mar 02, 2013 1:34 pm, edited 1 time in total.

The odds of getting heads and tails never changes, regardless of previous tosses.

If you have tossed 5 tails, you have already beaten odds of 1/32 (3.1%) To toss another tails would only halve your odds to 1.55%

Another way of looking at it is that, although when you started tossing you had a 3.1% chance of getting 5 tails. Having tossed them though, the chance that you did toss 5 tails is now 100%. Because it's in the past, we know the result 100%. Therefore, the next toss is still only 50/50.

If you have tossed 5 tails, you have already beaten odds of 1/32 (3.1%) To toss another tails would only halve your odds to 1.55%

Another way of looking at it is that, although when you started tossing you had a 3.1% chance of getting 5 tails. Having tossed them though, the chance that you did toss 5 tails is now 100%. Because it's in the past, we know the result 100%. Therefore, the next toss is still only 50/50.

In the test, 13 people were asked to flip a coin 300 times, trying to get as many heads as possible. All 13 participants got more heads than tails. Seven out of the thirteen had statistically significant margins of heads over tails (meaning almost certainly not a matter of chance). The highest was one individual had 68% of the coin flips land heads. In other words, a coin toss isn't particularly random.http://www.techdirt.com/articles/200912 ... 7292.shtml

I know from personal eperience that I can make a coin come out ether heads or tails...Most of the time.

If you have tossed 5 tails, you have already beaten odds of 1/32 (3.1%) To toss another tails would only halve your odds to 1.55%

Another way of looking at it is that, although when you started tossing you had a 3.1% chance of getting 5 tails. Having tossed them though, the chance that you did toss 5 tails is now 100%. Because it's in the past, we know the result 100%. Therefore, the next toss is still only 50/50.

In the test, 13 people were asked to flip a coin 300 times, trying to get as many heads as possible. All 13 participants got more heads than tails. Seven out of the thirteen had statistically significant margins of heads over tails (meaning almost certainly not a matter of chance). The highest was one individual had 68% of the coin flips land heads. In other words, a coin toss isn't particularly random.http://www.techdirt.com/articles/200912 ... 7292.shtml

I know from personal eperience that I can make a coin come out ether heads or tails...Most of the time.

Its indeed true, that you can make a head or tails appear more wich certain techiniques, but then lets go with somethign else. I have 2 balls a red and a white one, I put them both under a cup at random, you cant see through the cup, neither did you see me put the balls under the cup!

Now you can remove one of the caps, youll have to find the white ball to win!

Your odds to find the white one are 50%.

You fail the first time, second time you try now still 50% on finding it. THough in the sequence you failing twice has odds off 1/2^2, aka 1/4=0,25 aka 25%.

A couple of points .. and I realize the 100 heads results in a row example is just to illustrate the point that an independent toss of a coin has the same probability each and every time it is tossed.

I would contend there is a high probability that the coin is fixed, however. If asked to bet on the next flip, I would bet heads again because it seems probable that the coin is not true. Again, I know what concept was being taught there ...... just saying.

And another little tidbit ... most coins are not fifty - fifty propositions. Generally ( but not always ) the "heads" side of a coin is slightly heavier than a "tails" side of a coin and thus a tails result is ever so slightly more probable than a heads result. It obviously depends on the coin as to which side is heavier but the heavier side of a coin will generally face down a very very very slight majority of the flips.